Question
'A' operates at 75% of 'B's efficiency, while 'C'
functions at 125% of 'B's efficiency. Initially, 'A' and 'B' teamed up for the job and worked for 10 days before leaving. 'C' then took over and finished the remaining tasks by working solo for another 10 days. Determine how long it would take for 'A' and 'C' to jointly complete 80% of the work.Solution
Let the efficiency of 'B' be '4y' units/day So, efficiency of 'A' = 0.75 X 4y = '3y' units/day And, efficiency of 'C' = 1.25 X 4y = '5y' units/day Work done by 'A' and 'B' in 10 days = (3y + 4y) X 10 = '70y' units Work done by 'C' = 5y X 10 = '50y' units So, total work = 70y + 50y = '120y' Therefore, required time = (0.8 X 120y) ÷ (3y + 5y) = (96/8) = 12 days
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